For a parametric curve, dy/dx is (dy/dt) ÷ (dx/dt). The second derivative is the derivative of dy/dx with respect to t, divided again by dx/dt. Speed and arc length both use √((dx/dt)² + (dy/dt)²), Pythagoras applied to an infinitesimal step.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Why it works
The chain rule gives dy/dt = (dy/dx)(dx/dt). Rearranging isolates dy/dx without ever eliminating t. Check: x = t², y = t³ is y = x^(3/2), with y′ = (3/2)√x = (3/2)t. It agrees.
The second derivative
d²y/dx² = (d/dt of dy/dx) ÷ (dx/dt). For the curve above: dy/dx = 3t/2, its t-derivative is 3/2, divided by 2t gives 3/(4t). Dividing only once, getting 3/2, is the standard error.
Reading the second derivative
3/(4t) is positive for t > 0: concave up there. Negative for t < 0: concave down. The Cartesian check: y = x^(3/2) has y″ = (3/4)x^(−1/2) = 3/(4t). Same.
Vertical and horizontal tangents
Vertical where dx/dt = 0 and dy/dt ≠ 0. Horizontal the reverse. At t = 0 both are zero here: a cusp, and the formula says nothing. Check dx/dt before dividing.
Arc length
L = ∫ √((dx/dt)² + (dy/dt)²) dt between the t limits. For x = t², y = t³ from t = 0 to 1: ∫₀¹ √(4t² + 9t⁴) dt = ∫₀¹ t√(4 + 9t²) dt. Substituting u = 4 + 9t²: (1/18)(⅔)(13^(3/2) − 8) ≈ 1.44.
Speed
That same square root is the speed of the particle at time t: √(4t² + 9t⁴). At t = 1: √13 ≈ 3.61. Integrating speed over time gives distance traveled, which is the arc length.
Step 2: Try It Yourself
Tap and try it out.
x = t², y = t³. Find d²y/dx², and the arc length from t = 0 to t = 1
- 1dy/dx = 3t²/(2t) = 3t/2the first derivative
Step 0 of 5
- Point(2, 4)
- Slope of the tangent4
Step 3: In Real Life
Miles from a GPS track
From x(t) and y(t), the speed is √(x′² + y′²) and the distance is its integral: the arc length. That is how a phone turns a GPS track into miles run.
Step 4: Watch an Example
One step at a time.
Watch Rosa Find a Parametric Slope
Rosa has x = t² and y = t³, and needs dy/dx at t = 2.
- Step 1
Differentiating gives dx/dt = 2t and dy/dt = 3t², so dy/dx = 3t² ÷ 2t = 3t ÷ 2.
Step 5: Your Turn
Practice makes it stick.
The Slope
Problem 1 of 2
dy/dt = 12 and dx/dt = 4. What is dy/dx?
The Speed
Problem 2 of 2
dx/dt = 3 and dy/dt = 4. What is the speed?
Differentiate Parametrically
1 of 8
dy/dt = 10, dx/dt = 5. What is dy/dx?
2 of 8
dx/dt = 6, dy/dt = 8. What is the speed?
3 of 8
x = t², so dx/dt = 2t. What is dx/dt at t = 5?
4 of 8
y = t³, so dy/dt = 3t². What is dy/dt at t = 2?
5 of 8
dx/dt = 0 and dy/dt = 5. Is the tangent vertical or horizontal?
6 of 8
dx/dt = 5 and dy/dt = 0. Vertical or horizontal?
7 of 8
Put the process for the second derivative in order.
- 1Differentiate that expression with respect to t.
- 2Divide the result by dx/dt again.
- 3Simplify to get the second derivative.
- 4Compute dy/dx as the ratio of the two rates.
8 of 8
dx/dt = 5 and dy/dt = 12. What is the speed?
Step 6: Quick Check
Show what you know.
Question 1 of 2
dy/dt = 18 and dx/dt = 6. What is dy/dx?
Question 2 of 2
What is the common error when finding the second derivative?
What You Learned
- dy/dx is (dy/dt) ÷ (dx/dt).
- The second derivative divides by dx/dt a second time.
- Speed and arc length both use √((dx/dt)² + (dy/dt)²).
- Check dx/dt before dividing: where it is zero the tangent is vertical.